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2022 AMC 12A Problem 15

Problem 15 of 25IntermediateAlgebraGeometry

The roots of the polynomial 10x3−39x2+29x−610x^3-39x^2+29x-6 are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by 22 units. What is the volume of the new box?

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Solution

Let the roots be r,s,t.r,s,t. By Vieta’s formulas, r+s+t=3910,r+s+t=\dfrac{39}{10}, rs+rt+st=2910,rs+rt+st=\dfrac{29}{10}, and rst=610=35.rst=\dfrac{6}{10}=\dfrac35. The new volume is (r+2)(s+2)(t+2)=rst+2(rs+rt+st)+4(r+s+t)+8=35+5810+15610+8=30. \begin{gathered} (r+2)(s+2)(t+2) \\ =rst+2(rs+rt+st) \\ \quad {}+4(r+s+t)+8 \\ =\frac35+\frac{58}{10} \\ \quad {}+\frac{156}{10}+8 \\ =30. \end{gathered} Thus, the correct answer is D.
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Tagged: Vieta’s Formulas · volume

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